The bootstrap

Reference: Data Science Chapter 5

From the New England Journal of Medicine in 2006:

We randomly assigned patients with resectable adenocarcinoma of the stomach, esophagogastric junction, or lower esophagus to either perioperative chemotherapy and surgery (250 patients) or surgery alone (253 patients)…. With a median follow-up of four years, 149 patients in the perioperative-chemotherapy group and 170 in the surgery group had died. As compared with the surgery group, the perioperative-chemotherapy group had a higher likelihood of overall survival (five-year survival rate, 36 percent vs. 23 percent).

Conclusion:

• Chemotherapy patients are 13% more likely to survive past 5 years.
  

Conclusion:

• Chemotherapy patients are 13% more likely to survive past 5 years.
  

Not so fast! In statistics, we ask “what if?” a lot:

• What if the randomization of patients just happened, by chance, to assign more of the healthier patients to the chemo group?
  
• Or what if the physicians running the trial had enrolled a different sample of patients from the same clinical population?
  

Conclusion:

• Chemotherapy patients are 13% more likely to survive past 5 years.
  

Always remember two basic facts about samples:

• All numbers are wrong: any quantity derived from a sample is just a guess of the corresponding population-level quantity.
  
• A guess is useless without an error bar: an estimate of how wrong we expect the guess to be.
  

Conclusion:

• Chemotherapy patients are 13% \( \pm \) ? more likely to survive past 5 years, with ??% confidence.
  

By “quantifying uncertainty,” we mean filling in the blanks.

In stats, we equate trustworthiness with stability:

• If our data had been different merely due to chance, would our answer have been different, too?
• Or would the answer have been stable, even with different data?
  

\[ \begin{array}{r} \mbox{Confidence in} \\ \mbox{your estimates} \\ \end{array} \iff \begin{array}{l} \mbox{Stability of those estimates} \\ \mbox{under the influence of chance} \\ \end{array} \]

For example:

• If doctors had taken a different sample of 503 cancer patients and gotten a drastically different estimate of the new treatment's effect, then the original estimate isn't very trustworthy.
• If, on the other hand, pretty much any sample of 503 patients would have led to the same estimates, then their answer for this particular subset of 503 is probably accurate.
  

Let's work through a thought experiment…

Imagine Andrey Kolmorogov on four-day fishing trip.

• The lake is home to a very large population of fish of varying size and weight.
  
• On each day, Kolmorogov takes a random sample of size \( N=15 \) from this population—that is, he catches (and releases) 15 fish.
  
• He records the weight and approximate volume of each fish.
  
• He uses each day's catch to compute a different estimate of the volume–weight relationship for all fish in the lake.
  

Suppose we are trying to estimate some population-level quantity \( \theta \): the parameter of interest.

So we take a sample from the population: \( X_1, X_2, \ldots, X_N \).

We use the data to form an estimate \( \hat{\theta}_N \) of the parameter. Key insight: \( \hat{\theta}_N \) is a random variable.

Suppose we are trying to estimate some population-level quantity \( \theta \): the parameter of interest.

So we take a sample from the population: \( X_1, X_2, \ldots, X_N \).

We use the data to form an estimate \( \hat{\theta}_N \) of the parameter. Key insight: \( \hat{\theta}_N \) is a random variable.

Now imagine repeating this process thousands of times! Since \( \hat{\theta}_N \) is a random variable, it has a probability distribution.

Estimator: any method for estimating the value of a parameter (e.g. sample mean, sample proportion, slope of OLS line, etc).

Sampling distribution: the probability distribution of an estimator \( \hat{\theta}_N \) under repeated samples of size \( N \).

Bias: Let \( \bar{\theta}_N = E(\hat{\theta}_N) \) be the mean of the sampling distribution. The bias of \( \hat{\theta}_N \) is \( (\bar{\theta}_N - \theta) \): the difference between the average answer and the truth.

Unbiased estimator: \( (\bar{\theta}_N - \theta) = 0 \).

Standard error: the standard deviation of an estimator's sampling distribution:

\[ \begin{aligned} \mbox{se}(\hat{\theta}_N) &= \sqrt{ \mbox{var}(\hat{\theta}_N) } \\ &= \sqrt{ E[ (\hat{\theta}_N - \bar{\theta}_N )^2] } \\ &= \mbox{Typical deviation of $\hat{\theta}_N$ from its average} \end{aligned} \]

“If I were to take repeated samples from the population and use this estimator for every sample, how much does the answer vary, on average?”

If an estimator is unbiased, then \( \bar{\theta}_N = \theta \), so

\[ \begin{aligned} \mbox{se}(\hat{\theta}_N) &= \sqrt{ E[ (\hat{\theta}_N - \bar{\theta}_N )^2] } \\ &= \sqrt{ E[ (\hat{\theta}_N - \theta )^2] } \\ &= \mbox{Typical deviation of $\hat{\theta}_N$ from the truth} \end{aligned} \]

“If I were to take repeated samples from the population and use this estimator for every sample, how big of an error do I make, on average?”

This is why doctors and lawyers are buying “farmhouses.”

Chip and Joanna lifestyle item #1: the farmhouse sink

• On average across many weeks of manufacturing, the fancy sink has width equal to 30".
  
• But individual sinks vary from the average by about 0.5", due to manufacturing variability.
  
• So I expect that my specific sink will be somewhere in the vicinity of 30" \( \pm \) 0.5".
  

Don't make any lifestyle choices that require greater precision!

• On average across many samples, my estimator \( \hat{\theta}_N \) is equal to the right answer (\( \theta \)).
  
• But individual estimates vary from the average by about \( \mbox{se}(\hat{\theta}_N) \), due to sampling variability.
  
• So I expect that the right answer is somewhere in the vicinity of \( \hat{\theta}_N \pm \mbox{se}(\hat{\theta}_N) \).
  

Don't reach any scientific conclusions that require greater precision!

But there's a problem here…

• Knowing the standard error requires knowing what happens across many separate samples.
  
• But we've only got our one sample!
  
• So how can we ever calculate the standard error?
  

Two roads diverged in a yellow wood
And sorry I could not travel both
And be one traveler, long I stood
And looked down one as far as I could
To where it bent in the undergrowth…

–Robert Frost, The Road Not Taken, 1916

Quantifying our uncertainty would seem to require knowing all the roads not taken—an impossible task.

Problem: we can't take repeated samples of size \( N \) from the population, to see how our estimate changes across samples.

Seemingly hacky solution: take repeated samples of size \( N \), with replacement, from the sample itself, and see how our estimate changes across samples. This is something we can easily simulate on a computer.

Basically, we pretend that our sample is the whole population and we charge ahead! This is called bootstrap resampling, or just bootstrapping.

Bootstrapped sample 1

Bootstrapped sample 2

Bootstrapped sample 3

• Each bootstrapped sample has its own pattern of duplicates and omissions from the original sample.
  
• These duplicates and omissions create variability in \( \hat{\theta} \) from one bootstrapped sample to the next.
  
• This variability mimics the true sampling variability you'd expect to see across real repeated samples from the population.
  

• Start with your original sample \( S = \{X_1, \ldots, X_N\} \) and original estimate \( \hat{\theta}_N \).

• For \( b=1,...,B \):

  1. Take a bootstrapped sample \( S^{(b)} = \{ X_1^{(b)}, \ldots, X_N^{(b)} \} \)
     
  2. Use \( S^{(b)} \) to re-form the estimate \( \hat{\theta}_N^{(b)} \).

• Result: a set of \( B \) different estimates \( \hat{\theta}_N^{(1)}, \hat{\theta}_N^{(b)}, \ldots, \hat{\theta}_N^{(B)} \) that approximate the sampling distribution of \( \hat{\theta}_N \).

Calculate the bootstrapped standard error as the standard deviation of the bootstrapped estimates:

\[ \hat{se}(\hat{\theta}_N) = \mbox{std dev}\left( \hat{\theta}_N^{(1)}, \hat{\theta}_N^{(b)}, \ldots, \hat{\theta}_N^{(B)} \right) \]

This isn't the true standard error, but it's often a good approximation!

Let's dig in to some R code and data: creatinine_bootstrap.R and creatinine.csv (both on class website).

We'll bootstrap two estimators:

• the sample mean
  
• the OLS estimate of a slope
  

Informally, an interval estimate is a range of plausible values for the parameter of interest. For example:

• Go out some multiple \( k \) of the bootstrapped standard error from your estimate:
  

\[ \theta \in \hat{\theta}_N \pm k \cdot \hat{se} (\hat{\theta}_N) \]

• Use the quantiles (e.g. the 2.5 and 97.5 percentiles) of the bootstrapped sampling distribution, to cover a large fraction (e.g. 95%) of the bootstrapped estimates:

\[ \theta \in (q_{2.5}, q_{97.5}) \]

We'd like to be able to associate a confidence level with an interval estimate like this. How?

If an interval estimate satisfies the frequentist coverage principle, we call it a confidence interval:

Frequentist coverage principle: If you were to analyze one data set after another for the rest of your life, and you were to quote X% confidence intervals for every estimate you made, those intervals should cover their corresponding true values at least X% of the time. Here X can be any number between 0 and 100.

An interval estimate takes the form \( \hat{I}_N = [\hat{L}_N, \hat{U}_N] \). Just like a point estimate \( \hat{\theta}_N \), the interval estimate is a random variable, because its endpoints are functions of a random sample.

We say that \( [\hat{L}_N, \hat{U}_N] \) is a confidence interval at coverage level \( 1-\alpha \) if, for every \( \theta \),

\[ P_{\theta} \left( \theta \in [\hat{L}_N, \hat{U}_N] \right) \geq 1- \alpha \, , \]

where \( P_{\theta} \) is the probability distribution of the data, assuming that the true parameter is equal to \( \theta \).

The key statement here can be one of the most confusing in all of statistics:

\[ P_{\theta} \left( \theta \in [\hat{L}_N, \hat{U}_N] \right) \geq 1- \alpha \, , \]

Three questions to ask yourself:

• What is fixed?
  
• What is random?
  
• What is the source of this randomness?
  

So recall our two methods of generating an interval estimate using the bootstrap:

• The standard error method: \( \theta \in \hat{\theta}_N \pm z^{\star} \cdot \hat{se} (\hat{\theta}_N) \), where \( z^{\star} \) is a pre-specified quantile of the normal distribution.
  
• The quantile method (e.g. the 2.5 and 97.5 percentiles of the bootstrapped sampling distribution)
  

The obvious question is: do these interval estimates satisfy the frequentist coverage principle?

The answer is: not always, but often!

In lots of common situations, both forms of bootstrapped interval estimate approximately satisfy the coverage requirement:

\[ P_{\theta} \left( \theta \in [\hat{L}_N, \hat{U}_N] \right) \approx 1- \alpha \, , \]

And the approximation gets better with larger sample sizes. That is, as \( N \) gets large,

\[ P_{\theta} \left( \theta \in [\hat{L}_N, \hat{U}_N] \right) \to 1-\alpha \]

The math here is super hairy; we won't go into it. (Google “empirical process theory” if want to learn and you've got a year or two to spare…)

But we can run a sanity check through Monte Carlo simulation!

Let's revisit our thought experiment about fishing…

Let's go on a 100 fishing trips. On each trip:

• we catch a sample of \( N=30 \) fish
  
• we run OLS on our sample to estimate \( \beta_1 \): the slope of the weight-vs.-volume line
  
• we bootstrap our sample to get a 80% confidence interval for the slope.
  

Because we know the slope of the true line (\( \beta_1 = 4.25 \)), we can check whether each bootstrapped confidence interval contains the true value. About 80% of them should!

• 100 different samples
• 100 different 80% confidence intervals
• 83 of them cover the truth
  

• Any estimator \( \hat{\theta}_N \) is a random variable.
  
• It's probability distribution is called the sampling distribution.
  
• The sampling distribution describes the results of a thought experiment: what would happen if we took lots and lots of samples, each of size \( N \), and tracked how much our estimate changed?
  

• The standard error is the standard deviation of the sampling distribution.
  
• Roughly speaking, it answers the question: how far off do I expect my estimate to be from the truth?
  
• A practical way of estimating the standard error is by bootstrapping: repeatedly re-sampling with replacement from the original sample, and re-calculating the estimate each time.
  

• From the bootstrapped sampling distribution, we can get an interval estimate for the parameter of interest (using the standard-error method or the quantile method).
  
• Both methods approximately satisfy the frequentist coverage principle: under repeated sampling, they contain the true value roughly the correct percentage of the time.
  
• I tend to use the quantile method because it's pretty intuitive!
  

The essential idea of the bootstrap is to simulate synthetic data sets by resampling from the original sample. This is often called the nonparametric bootstrap.

A common variation is called the parametric bootstrap: simulate synthetic data sets by simulating from a fitted parametric model.

See predimed_bootstrap.R!